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12345 [234]
3 years ago
11

What is the 1/2 half life of Cesium 137 in years?

Chemistry
2 answers:
blondinia [14]3 years ago
8 0

Answer:

About 30.17 years

Explanation:

About 94.6% decays by beta emission to a metastable nuclear isomer of barium: barium-137m . The remainder directly populates the ground state of barium-137, which is stable.

svetoff [14.1K]3 years ago
4 0

Answer:

about 30.17 years

Explanation:

Caesium-137 has a half-life of about 30.17 years. About 94.6% decays by beta emission to a metastable nuclear isomer of barium: barium-137m (137mBa, Ba-137m). The remainder directly populates the ground state of barium-137, which is stable.

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If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr
Fynjy0 [20]
Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.
7 0
3 years ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
If the distance between a point charge and a neutral atom and is multiplied by a factor of 5, by what factor does the force on t
dexar [7]

Given :

The distance between a point charge and a neutral atom and is multiplied by a factor of 5.

To Find :

By what factor does the force on the neutral atom by the point charge change.

Solution :

We know, electrostatic force between two object is directly proportional to product of charge and inversely proportional to distance between them.

Now, charge in neutral atom is 0 C.

So, the electrostatic force between two of them is also 0 N.

Therefore, by changing distance between the charge the forces did no change ( it remains zero).

3 0
3 years ago
3. Challenge: Select Distance again from the Conversion menu. Use the conversion factors on the tiles to
Gnom [1K]

Answer: 1 liter

Explanation:

3 0
3 years ago
Calculate the volume of a sample of aluminum that has a mass of 3.057 kg. The density of aluminum is 2.70 g/cm3 (taking into con
Misha Larkins [42]

Answer:

v = 1130 cm³

Explanation:

Given data:

Volume of sample = ?

Mass of Al sample = 3.057 Kg (3.057 Kg× 1000g/1 Kg = 3057g)

Density of Al sample = 2.70 g/cm³

Solution:

Formula:

d = m/v

d = density

m = mass

v= volume

by putting values

2.70 g/cm³ = 3057g /v

v = 3057g /2.70 g/cm³

v = 1130 cm³

3 0
3 years ago
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